The RD Sharma Class 12 Chapter 3 Solutions are the best to buy. The chapters and exercises in RD Sharma Class 12 may be difficult for a student to grasp. They will find it easier with RD Sharma Class 12th Exercise 3.14. Students can now use these solutions to solve any Relationship problem. The questions and answers are written in a way that students can easily understand. These solutions lay the groundwork for students' understanding of mathematics. The concepts are now easily understood by the students.
Many students find mathematics a difficult subject. These solutions will make students happy because they will learn all of the concepts quickly and easily. With these solutions, the approach to learning and understanding topics will be forever altered. RD Sharma Solutions is here to ensure that every student excels in mathematics, so the book includes various tricks and tips.
RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise
Inverse Trigonometric Functions Excercise: 3.14
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download E-bookInverse Trigonometric Functions Exercise 3.14 Question 1(i)
Answer: $\frac{-7}{17}$ Hints: First we will solve for
$2 \tan ^{-1} \frac{1}{5}$ and we use the formula
$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right),-1<x<1$ Given: $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$ Explanation: $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$ ..............(1) Let us first solve for
$2 \tan ^{-1} \frac{1}{5}$ $2 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left\{\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right\}$ $\left[\because-1<\frac{1}{5}<1\right]$ $=\tan ^{-1}\left\{\frac{\frac{2}{5}}{1-\left(\frac{1}{25}\right)}\right\}$ $=\tan ^{-1}\left\{\frac{\frac{2}{5}}{\left(\frac{24}{25}\right)}\right\}$ $=\tan ^{-1}\left\{\frac{2}{5} \times \frac{25}{24}\right\}$ $2 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left(\frac{5}{12}\right)$ ................(2)
Now, we know that
$\tan \frac{\pi}{4}=1$ $\frac{\pi}{4}=\tan ^{-1}(1)$ Now from equation (1), (2) and (3)
We get
$\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$ $=\tan \left\{\tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1}(1)\right\}$ $=\tan \left\{\tan ^{-1}\left(\frac{\frac{5}{12}-1}{1+\frac{5}{12} \times 1}\right)\right\}$ $\left[\begin{array}{l} \because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right) \\ \text { forxy }>-1 \end{array}\right]$ $=\tan \left\{\tan ^{-1}\left(\frac{\frac{-7}{12}}{\frac{17}{12}}\right)\right\}$ $=\tan \left\{\tan ^{-1}\left(\frac{-7}{12} \times \frac{12}{17}\right)\right\}$ $=\tan \left\{\tan ^{-1}\left(\frac{-7}{17}\right)\right\}$ $\left[\because \tan \left(\tan ^{-1} \theta\right)=\theta\right]$ $=\frac{-7}{17}$
Answer: $\frac{4-\sqrt{7}}{3}$ Hints: First we will convert $\sin ^{-1}\left(\frac{3}{4}\right)$ into $\tan ^{-1}$ Given: $\tan \left(\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)\right)$ Explanation: Let$\sin ^{-1}\left(\frac{3}{4}\right)=\theta$ ......(1)$\sin \theta=\frac{3}{4}=\frac{A B}{A C}$ In$\triangle A B C$ $A B^{2}+B C^{2}=A C^{2}$ (By using Pythagoras theorem)$\begin{aligned} &3^{2}+B C^{2}=4^{2} \\ &9+B C^{2}=16 \\ &B C^{2}=16-9 \\ &B C^{2}=7 \\ &B C=\sqrt{7} \end{aligned}$ $\begin{aligned} &\cos \theta=\frac{B C}{A C} \\ &\cos \theta=\frac{\sqrt{7}}{4} \end{aligned}$ Now$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$ $\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\frac{\sqrt{7}}{4}}{1+\frac{\sqrt{7}}{4}}}$ $\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{\frac{4-\sqrt{7}}{4}}{\frac{4+\sqrt{7}}{4}}}$ $\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}$ $\frac{\theta}{2}=\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)$ $\frac{1}{2} \sin ^{-1} \frac{3}{4}=\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)$ from equation (1) Now, $\tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right)$ $=\tan \left(\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)\right)$ from equation (2)$=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}$ On rationalizing we get,$\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$ $=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}} \times \frac{4-\sqrt{7}}{4-\sqrt{7}}}$ $=\sqrt{\frac{(4-\sqrt{7})^{2}}{16-7}}$ $\begin{aligned} &=\sqrt{\frac{(4-\sqrt{7})^{2}}{9}} \\ &=\frac{(4-\sqrt{7})}{3} \end{aligned}$
Answer: $\frac{1}{\sqrt{10}}$ Hint: First we will convert $\cos ^{-1} \frac{4}{5}$ into $\sin ^{-1}$ Given: $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$ Explanation: Let$\cos ^{-1} \frac{4}{5}=\theta$ $\cos \theta=\frac{4}{5}$ we know that$\begin{aligned} &\cos \theta=1-2 \sin ^{2} \frac{\theta}{2} \\ &2 \sin ^{2} \frac{\theta}{2}=1-\cos \theta \\ &\sin ^{2} \frac{\theta}{2}=\frac{1-\cos \theta}{2} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{2}} \end{aligned}$ $\begin{aligned} &\sin \frac{\theta}{2}=\sqrt{\frac{1-\frac{4}{5}}{2}} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{\frac{1}{5}}{2}} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{1}{10}} \\ &\sin \frac{\theta}{2}=\frac{1}{\sqrt{10}} \\ &\frac{\theta}{2}=\sin ^{-1} \frac{1}{\sqrt{10}} \end{aligned}$ from equation (1)$\frac{1}{2} \cos ^{-1} \frac{4}{5}=\sin ^{-1} \frac{1}{\sqrt{10}}$ ......(2) Now$\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$ from equation (2)$=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)$ $=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)$ $\left[\because \sin \left[\sin ^{-1}(\theta)\right]=\theta\right]$ $=\frac{1}{\sqrt{10}}$
Answer: $\frac{37}{26}$ Hints: we will convert $2 \tan ^{-1} \frac{2}{3}$ into $\sin ^{-1}$ and $\tan ^{-1} \sqrt{3}$ into $\cos ^{-1}$ Given: $\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$ Solution: $\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$ ..................(1) First we will solve for $2 \tan ^{-1} \frac{2}{3}$ We know that $2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \quad-1<x<1$ $\tan ^{-1} \frac{2}{3}=\sin ^{-1}\left(\frac{2 \times \frac{2}{3}}{1+\left(\frac{2}{3}\right)^{2}}\right)$ $-1<\frac{2}{3}<1$ $\begin{aligned} &=\sin ^{-1}\left(\frac{\frac{4}{3}}{1+\frac{4}{9}}\right) \\ &=\sin ^{-1}\left(\frac{\frac{4}{3}}{\frac{13}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{3} \times \frac{9}{13}\right) \end{aligned}$ $2 \tan ^{-1} \frac{2}{3}=\sin ^{-1}\left(\frac{12}{13}\right)$ .................(2) Now Let $\tan ^{-1} \sqrt{3}=\theta$ .................(3)$\tan \theta=\frac{\sqrt{3}}{1}=\frac{P}{B}$ By Pythagoras theorem:$\begin{aligned} &H^{2}=P^{2}+B^{2} \\ &=\sqrt{3}^{2}+1^{2} \\ &=3+1 \\ &=4 \\ &H=\sqrt{4} \\ &H=2 \\ &\cos \theta=\frac{B}{H} \\ &\cos \theta=\frac{1}{2} \end{aligned}$ $\theta=\cos ^{-1}\left(\frac{1}{2}\right)$ from equation.........(2)$\tan ^{-1} \sqrt{3}=\cos ^{-1}\left(\frac{1}{2}\right)$ ..................(4) Now from equation (1)$\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$ Putting the value of $2 \tan ^{-1} \frac{2}{3}$ and $\tan ^{-1} \sqrt{3}$ from equations (2) and (4) respectively$\begin{aligned} &=\sin \left(\sin ^{-1} \frac{12}{13}\right)+\cos \left(\cos ^{-1} \frac{1}{2}\right) \\ &=\frac{12}{13}+\frac{1}{2} \\ &=\frac{24+13}{26} \\ &=\frac{37}{26} \end{aligned}$
Answer: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$ Hints: First we will convert $\sin ^{-1} \frac{3}{5}$ into $\tan ^{-1}$ and after that we use the formula of $2 \tan ^{-1} x$ $2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \quad-1<x<1$ Given: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$ Explanation: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$ First we will solve for $\sin ^{-1} \frac{3}{5}$ let $\sin ^{-1} \frac{3}{5}=\theta$ ........................(1)$\sin \theta=\frac{3}{5}=\frac{P}{H}$ By Pythagoras theorem$\begin{aligned} &H^{2}=P^{2}+B^{2} \\ &5^{2}=3^{2}+B^{2} \\ &25=9+B^{2} \\ &B^{2}=25-9 \\ &B^{2}=16 \\ &B=\sqrt{16} \\ &B=4 \end{aligned}$ $\tan \theta=\frac{P}{B}$ $\tan \theta=\frac{3}{4} \quad \theta=\tan ^{-1} \frac{3}{4}$ From Equation (1)$\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$ ..............(2) Now L.H.S:$2 \sin ^{-1} \frac{3}{5}$ $=2 \tan ^{-1} \frac{3}{4}$ from equation (2)$=\tan ^{-1}\left[\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right]$ $\left[-1<\frac{3}{4}<1\right.$ $=\tan ^{-1}\left[\frac{\frac{3}{2}}{1-\frac{9}{16}}\right]$ $=\tan ^{-1}\left[\frac{\frac{3}{2}}{\frac{7}{16}}\right]$ $\begin{aligned} &=\tan ^{-1}\left[\frac{3}{2} \times \frac{16}{7}\right] \\ &=\tan ^{-1}\left[\frac{24}{7}\right] \\ &=R \cdot H . S \end{aligned}$ Hence $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$
Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 2
Answer: $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}$ Hints: First we will solve for
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$ then we will convert it into
$\cos ^{-1} \text { and } \sin ^{-1}$ respectively.
Given: $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}$ First we will solve for
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$ $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4}\right)\left(\frac{2}{9}\right)}\right)$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{9+8}{36}}{1-\frac{1}{18}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right) \\ &=\tan ^{-1}\left(\frac{17}{36} \times \frac{18}{17}\right) \end{aligned}$ $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\tan ^{-1} \frac{1}{2}$ $=\frac{1}{2} \times 2 \tan ^{-1} \frac{1}{2}$ (On multiplying and dividing by 2)
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \times \cos ^{-1}\left(\frac{1-\left(\frac{1}{2}\right)^{2}}{1+\left(\frac{1}{2}\right)^{2}}\right) \quad\left(\begin{array}{l} \because 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\ 0 \leq x<\infty \end{array}\right)$ $\begin{aligned} &=\frac{1}{2} \cos ^{-1}\left(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}\right) \\ &=\frac{1}{2} \cos ^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right) \end{aligned}$ $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$ .......................(1)
Now,let
$\cos ^{-1} \frac{3}{5}=\theta$ ....................(2)
$\begin{aligned} &\cos \theta=\frac{3}{5} \\ &\therefore \sin \theta=\sqrt{1-\cos ^{2} \theta} \end{aligned}$ $\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$ $\begin{aligned} \sin \theta &=\sqrt{1-\cos ^{2} \theta} \\ \sin \theta &=\sqrt{1-\left(\frac{3}{5}\right)^{2}} \\ &=\sqrt{1-\frac{9}{25}} \\ &=\sqrt{\frac{25-9}{25}} \\ &=\sqrt{\frac{16}{25}} \end{aligned}$ $\begin{aligned} &\sin \theta=\frac{4}{5} \\ &\theta=\sin ^{-1} \frac{4}{5} \end{aligned}$ from equation (2)
$\cos ^{-1} \frac{3}{5}=\sin ^{-1} \frac{4}{5}$ .........(3)
From equation (1) and (2) we get
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}$
Answer: $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$ Hint: First we will multiply in numerator and denominator by 2 in L.H.S so that we can use the formula of $2 \tan ^{-1} x$ Given: $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$ Explanation: L.H.S:$\tan ^{-1} \frac{2}{3}$ On multiply in numerator and denominator by 2.$=\frac{1}{2} \times 2 \tan ^{-1} \frac{2}{3}$ $=\frac{1}{2} \times \tan ^{-1}\left(\frac{2 \times \frac{2}{3}}{1-\left(\frac{2}{3}\right)^{2}}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{2}{3}<1 \end{array}\right]$ $\begin{aligned} &=\frac{1}{2} \times \tan ^{-1}\left(\frac{\frac{4}{3}}{1-\frac{4}{9}}\right) \\ &=\frac{1}{2} \times \tan ^{-1}\left(\frac{\frac{4}{3}}{\frac{5}{9}}\right) \\ &=\frac{1}{2} \times \tan ^{-1}\left(\frac{4}{3} \times \frac{9}{5}\right) \\ &=\frac{1}{2} \tan ^{-1}\left(\frac{12}{5}\right) \end{aligned}$ Hence it is proved.
Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 4
Answer: $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$ Hints: First we will use the formula of
$2 \tan ^{-1} x$ .
Given: $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$ Explanation: $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{1}{3}<1 \end{array}\right]$ L.H.S:
$\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}$ $=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right)$ $\begin{aligned} &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{\frac{2}{3}}{1-\frac{1}{9}}\right) \\ &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right) \\ &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right) \end{aligned}$ $\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{3}{28}<1 \end{array}\right]$ $=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{3}{4}\right)$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{3}{4}}{1-\left(\frac{1}{7}\right)\left(\frac{3}{4}\right)}\right) \\ &=\tan ^{-1}\left(\frac{\frac{4+21}{28}}{\frac{28-3}{28}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right) \\ &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\begin{array}{l} \because \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$ Hence it is proved that
$\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$ Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 5
Answer: $\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}$ Hints: First we will convert $2 \tan ^{-1}\left(\frac{1}{3}\right)$ in $\sin ^{-1}$ Given: $\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}$ Explanation: L.H.S:$\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)$ $\left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right]$ $=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{2 \times \frac{1}{3}}{1+\left(\frac{1}{3}\right)^{2}}\right)$ $\left[\begin{array}{c} -1 \leq x \leq 1 \\ -1 \leq \frac{1}{3} \leq 1 \end{array}\right]$ $\begin{aligned} &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{\frac{2}{3}}{1+\frac{1}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{\frac{2}{3}}{\frac{10}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{2}{3} \times \frac{9}{10}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{3}{5}\right) \end{aligned}$ $\left[\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right\}\right]$ $\text { [if } \left.x, y \geq 0 \text { and } x^{2}+y^{2} \leq 1\right]$ $\left[\begin{array}{l} \frac{4}{5}, \frac{3}{5} \geq 0,\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2} \\ \frac{16}{25}+\frac{9}{25}=\frac{25}{25}=1 \leq 1 \end{array}\right]$ $\begin{aligned} &=\sin ^{-1}\left\{\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right\} \\ &=\sin ^{-1}\left(\frac{4}{5} \sqrt{1-\frac{9}{25}}+\frac{3}{5} \sqrt{1-\frac{16}{25}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5} \sqrt{\frac{16}{25}}+\frac{3}{5} \sqrt{\frac{9}{25}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5} \times \frac{4}{5}+\frac{3}{5} \times \frac{3}{5}\right) \\ &=\sin ^{-1}\left(\frac{16}{25}+\frac{9}{25}\right) \\ &=\sin ^{-1}\left(\frac{25}{25}\right) \end{aligned}$ $\left[\begin{array}{l} \because \sin \frac{\pi}{2}=1 \\ \sin ^{-1}(1)=\frac{\pi}{2} \end{array}\right]$ $\begin{aligned} &=\sin ^{-1}(1) \\ &=\frac{\pi}{2} \end{aligned}$ Hence it is proved that$\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}$
Answer: $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$ Hint: First we will convert $\sin ^{-1}\left(\frac{3}{5}\right)$ into $\tan ^{-1}$ .Given: $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$ Explanation: Let us solve for $\sin ^{-1}\left(\frac{3}{5}\right)$ Let $\sin ^{-1}\left(\frac{3}{5}\right)=\theta$ ............(1)$\begin{aligned} &\sin \theta=\left(\frac{3}{5}\right)=\frac{P}{H} \\ &H^{2}=P^{2}+B^{2} \\ &5^{2}=3^{2}+B^{2} \\ &25=9+B^{2} \\ &B^{2}=25-9 \\ &B^{2}=16 \\ &B=\sqrt{16} \\ &B=4 \end{aligned}$ Now,$\begin{aligned} &\tan \theta=\frac{P}{B} \\ &\tan \theta=\frac{3}{4} \\ &\theta=\tan ^{-1}\left(\frac{3}{4}\right) \end{aligned}$ From equation (1)$\sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{3}{4}\right)$ .............(2) L.H.S:$2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$ from Equation (2)$=2 \tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$ $\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$ $=\tan ^{-1}\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$ $\left[\begin{array}{c} -1<x<1 \\ -1<\frac{3}{4}<1 \end{array}\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{3}{2}}{1-\frac{9}{16}}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{24}{7}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \end{aligned}$ $\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$ $=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right)$ $\left[\begin{array}{l} x \cdot y>-1 \\ \frac{24}{7} \times \frac{17}{31}=\frac{408}{217}>-1 \end{array}\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{408}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{744-119}{217}}{\frac{625}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right) \end{aligned}$ $\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$ $\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ Hence it is Prove that $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$
Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 7 . Answer:
$2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$ Hint: First we will solve for $2 \tan ^{-1} \frac{1}{5}$ Given: $2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$ Explanation: L.H.S:$2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}$ $=\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{1}{5}<1 \end{array}\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\left(\frac{1}{25}\right)}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{2}{5} \times \frac{25}{24}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \end{aligned}$ $\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{5}{12} \times \frac{1}{8}=\frac{5}{96}<1 \end{array}\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{5}{12}+\frac{1}{8}}{1-\frac{5}{12} \times \frac{1}{8}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{10+3}{24}}{1-\frac{5}{96}}\right) \end{aligned}$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{13}{24}}{\frac{91}{96}}\right) \\ &=\tan ^{-1}\left(\frac{13}{24} \times \frac{96}{91}\right) \\ &=\tan ^{-1}\left(\frac{4}{7}\right) \end{aligned}$ Hence it is proved that $2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$
Answer: $2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$ Hint: First we will solve for $2 \tan ^{-1} \frac{3}{4}$ Given: $2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$ Explanation: L.H.S:$2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{3}{4}<1 \end{array}\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{3}{2}}{1-\frac{9}{16}}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{24}{7}\right)-\tan ^{-1} \frac{17}{31} \end{aligned}$ $=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right)$ $\left[\begin{array}{l} \because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right), x y>-1 \\ \frac{24}{7} \times \frac{17}{31}>-1 \end{array}\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{744-119}{217}}{1+\frac{408}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right) \end{aligned}$ $\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$ Hence it is proved that$2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$
Answer: $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$ Hint: First we will solve for $2 \tan ^{-1} \frac{1}{2}$ Given: $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$ Explanation: $2 \tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \end{array}\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{1}{1-\left(\frac{1}{4}\right)}\right) \\ &=\tan ^{-1}\left(\frac{1}{3}{4}\right) \end{aligned}$ $2 \tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{4}{3}\right)$ ...................(1) L.H.S:$2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}$ From equation (1)$=\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$ $=\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}\right)$ $\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{4}{3} \times \frac{1}{7}=\frac{4}{21}<1 \end{array}\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right) \\ &=\tan ^{-1}\left(\frac{31}{21} \times \frac{21}{17}\right) \\ &=\tan ^{-1}\left(\frac{31}{17}\right) \end{aligned}$ Hence it is proved that $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$
Inverse Trigonomeric Functions Exercise 3.14 Question 2 Sub Question 10.
Answer: $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}$ Hints: First we will solve for
$4 \tan ^{-1} \frac{1}{5}$ Split into
$2 \times 2 \tan ^{-1} \frac{1}{5}$ and solve it
Given: $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}$ Explanation : Let us first solve
$4 \tan ^{-1} \frac{1}{5}$ $4 \tan ^{-1} \frac{1}{5}=2 \times 2 \tan ^{-1} \frac{1}{5}$ $=2 \tan ^{-1}\left(\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{1}{5}<1 \end{array}\right]$ $\begin{aligned} 4 \tan ^{-1} \frac{1}{5} &=2 \tan ^{-1}\left(\frac{\frac{2}{5}}{1-\frac{1}{25}}\right) \\ &=2 \tan ^{-1}\left(\frac{2}{5} \times \frac{25}{24}\right) \\ &=2 \tan ^{-1}\left(\frac{5}{12}\right) \end{aligned}$ $=\tan ^{-1}\left(\frac{2 \times \frac{5}{12}}{1-\left(\frac{5}{12}\right)^{2}}\right) \quad\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<\frac{5}{12}<1 \end{array}\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{5}{6}}{1-\frac{25}{144}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{119}{144}}\right) \\ &=\tan ^{-1}\left(\frac{5}{6} \times \frac{144}{119}\right) \end{aligned}$ $4 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left(\frac{120}{119}\right)$ ......(1)
Now
L.H.S:
$\begin{aligned} &4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239} \\ &=\tan ^{-1}\left(\frac{120}{119}\right)-\tan ^{-1}\left(\frac{1}{239}\right) \\ &=\tan ^{-1}\left(\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119} \times \frac{1}{239}}\right) \end{aligned}$ $\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{28680-119}{28441}}{\frac{28441+120}{28441}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{28561}{28441}}{\frac{28561}{28441}}\right) \end{aligned}$ $\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$ Hence it is proved that
$4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}$ Inverse Trigonomeric Functions Exercise 3.14 Question 3 .
Answer: $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}} \text { then } x=\frac{a-b}{1+a b}$ Hints: First we will convert whole L.H.S part in
$\tan ^{-1}$ Given:$\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$ Explanation: As we know that
$\begin{aligned} &2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}, 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ &2 \tan ^{1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}$ Now
$\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$ $\begin{aligned} &2 \tan ^{-1} a-2 \tan ^{-1} b=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2\left(\tan ^{-1} a-\tan ^{-1} b\right)=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2 \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2 \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x \\ &\frac{a-b}{1+a b}=\tan \left(\tan ^{-1} x\right) \\ &\frac{a-b}{1+a b}=x \\ &x=\frac{a-b}{1+a b} \end{aligned}$ Hence it is proved that
$x=\frac{a-b}{1+a b}$ Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 1 .
Answer: $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$ Hint: First we will convert
$\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)$ into
$\tan ^{-1}$ Given: $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$ Explanation: L.H.S:
$\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)$ $=\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$ $\left[\because \cot ^{-1} x=\tan ^{-1}\left(\frac{1}{x}\right)\right]$ $=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-\left(\frac{1-x^{2}}{2 x}\right)\left(\frac{2 x}{1-x^{2}}\right)}\right]$ $\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$ $\begin{aligned} &=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-1}\right] \\ &=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{0}\right] \end{aligned}$ $\left(\frac{a}{0}=\infty\right)$ $=\tan ^{-1}(\infty)$ $\left(\begin{array}{l} \because \tan \frac{\pi}{2}=\infty \\ \tan ^{-1}(\infty)=\frac{\pi}{2} \end{array}\right)$ $=\frac{\pi}{2}$ Hence it is Proved that
$\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$ Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 2.
Answer: $\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1$ Hint: First we will convert $\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$ into $\tan ^{-1}$ Given: $\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1$ Explanation: L.H.S:$\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}$ $=\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+2 \tan ^{-1} x\right\}$ $\left[\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x\right]$ $=\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\tan ^{-1} \frac{2 x}{1-x^{2}}\right\}$ $\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]$ $\begin{aligned} &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-\left(\frac{1-x^{2}}{2 x}\right)\left(\frac{2 x}{1-x^{2}}\right)}\right\} \\ &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-1}\right\} \\ &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{0}\right\} \end{aligned}$ $\left(\frac{a}{0}=\infty\right)$ $=\sin \left\{\tan ^{-1}(\infty)\right\}$ $\left(\begin{array}{l} \because \tan \frac{\pi}{2}=\infty \\ \tan ^{-1}(\infty)=\frac{\pi}{2} \end{array}\right)$ $\begin{aligned} &=\sin \left(\frac{\pi}{2}\right) \\ &=1 \end{aligned}$ Hence it is proved that $\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1$
Answer: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$ Hints: We will first convert $2 x \sqrt{1-x^{2}}$ into sinGiven: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$ $x=\sin \theta$ Let $\sqrt{1-x^{2}}=\sqrt{1-\sin ^{2} \theta}=\sqrt{\cos ^{2} \theta}=\cos \theta$ $\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \cos ^{2} \theta=1-\sin ^{2} \theta \end{array}\right]$ Now$\begin{aligned} &\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \\ &=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \\ &=\sin ^{-1}(2 \sin \theta \cos \theta) \end{aligned}$ $[2 \sin \theta \cos \theta=\sin 2 \theta]$ $\begin{aligned} &=2 \theta \\ &=2 \sin ^{-1} x \end{aligned}$ $\left[\begin{array}{l} \because \sin ^{-1}(\sin \theta)=\theta \\ x=\sin \theta \\ \theta=\sin ^{-1} x \end{array}\right]$ Hence it is proved that $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$
Answer: $\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x$ to prove $x=\frac{a+b}{1-a b}$ Hints: First we will convert L.H.S of the question in $\tan ^{-1}$ Given: $\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x$ Explanation: $\begin{aligned} &\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x \\ &\because 2 \tan ^{-1} a=\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right) \\ &\therefore \sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x \end{aligned}$ $\begin{aligned} &2 \tan ^{-1} a+2 \tan ^{-1} b=2 \tan ^{-1} x \\ &2\left(\tan ^{-1} a+\tan ^{-1} b\right)=2 \tan ^{-1} x \\ &\tan ^{-1} a+\tan ^{-1} b=\tan ^{-1} x \end{aligned}$ $\tan ^{-1}\left(\frac{a+b}{1-a b}\right)=\tan ^{-1} x$ $\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$ $\begin{aligned} &\frac{a+b}{1-a b}=\tan \left(\tan ^{-1} x\right) \\ &\frac{a+b}{1-a b}=x \end{aligned}$ Hence it is proved that $x=\frac{a+b}{1-a b}$
Inverse Trigonomeric Functions Exercise 3.14 Question 6.
Answer: $\pi$ Hint: First we will convert $\sin ^{-1} \frac{2 x}{1+x^{2}}$ into $\tan ^{-1}$ Given: $2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}}$ for $x \geq 1$ Explanation: $2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}}$ $\left[\because 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}\right]$ $\begin{aligned} &=2 \tan ^{-1} x+2 \tan ^{-1} x \\ &=4 \tan ^{-1} x \quad(x \geq 1) \end{aligned}$ Let $x=1$ $\left[\begin{array}{l} \because \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$ $\begin{aligned} &=4 \tan ^{-1}(1) \\ &=4 \times \frac{\pi}{4} \\ &=\pi \end{aligned}$ Hence the constant value for $x \geq 1$ is $\pi$
Inverse Trigonomeric Functions Exercise 3.14 Question 7 Sub Question 1.
Answer:
$\frac{\pi}{4}$ Given: Given that
$\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$ Hint: First we solve
$2 \sin ^{-1} \frac{1}{2}$ Since
$\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$ Solution: We have,
$\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$ $=\tan ^{-1}\left\{2 \cos \left(2 \times \frac{\pi}{6}\right)\right\}$ $\left[\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}\right]$ $\begin{aligned} &=\tan ^{-1}\left\{2 \cos \frac{\pi}{3}\right\} \\ &=\tan ^{-1}\left\{2 \times \frac{1}{2}\right\} \end{aligned}$ $\left[\cos \frac{\pi}{3}=\frac{1}{2}\right]$ $\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\tan ^{-1}(1)=\frac{\pi}{4}\right]$ This is the required solution.
Inverse Trigonomeric Functions Exercise 3.14 Question 7 Sub Question 2.
Answer: 0Given: $\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right),|x| \geq 1$ Hint: Since $\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}$ applying it. Solution: We have,$\begin{aligned} &\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right) \\ &=\cos \left(\frac{\pi}{2}\right) \end{aligned}$ $\left[\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}\right]$ =0 $\left[\cos \frac{\pi}{2}=0\right]$ Hence, $\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right)=0$
Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 1.
Answer: $x=\frac{-461}{9}$ Given: $\tan ^{-1}\left(\frac{1}{4}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$ Hint: Use formula
$\text { 1. } \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\text { 2. } 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$ Solution: We have,
$\tan ^{-1}\left(\frac{1}{4}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$ $\Rightarrow \tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$ $\left[2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$ $\Rightarrow \tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$ We know that
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{4}+\frac{5}{12}}{1-\frac{1}{4} \times \frac{5}{12}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{8}{12}}{\frac{43}{48}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{32}{43}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \end{aligned}$ $\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$ $\Rightarrow \tan ^{-1}\left(\frac{\frac{32}{43}+\frac{1}{6}}{1-\frac{32}{43} \times \frac{1}{6}}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$ $\left[\tan ^{-1} \frac{\pi}{4}=1\right]$ $\Rightarrow \tan ^{-1}\left(\frac{\frac{235}{258}}{\frac{226}{258}}\right)+\tan ^{-1}\left(\frac{1}{\mathrm{x}}\right)=\tan ^{-1}(1)$ $\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{235}{226}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\tan ^{-1}(1) \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{235}{226}+\frac{1}{x}}{1-\left(\frac{235}{226}\right) \times \frac{1}{x}}\right)=\tan ^{-1}(1), \frac{235}{226} x<1 \end{aligned}$ $\begin{aligned} &\Rightarrow \frac{235 x+226}{226 x-235}=1, x>\frac{235}{226} \\ &\Rightarrow 235 x+226=226 x-235, x>\frac{235}{226} \\ &\Rightarrow 235 x-226 x=-235-226 \\ &\Rightarrow 9 x=-461 \\ &\Rightarrow x=\frac{-461}{9} \end{aligned}$ Hence
$x=\frac{-461}{9}$ is required answer
Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 2.
Answer: $x=\frac{1}{\sqrt{3}}$ Given: $3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$ Hint: Using formula since,
$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\sin ^{-1} \frac{2 x}{1+x^{2}}=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ Solution: We have,
$3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$ We know that
$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\sin ^{-1} \frac{2 x}{1+x^{2}}=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ $\begin{aligned} &\Rightarrow 3\left(2 \tan ^{-1} x\right)-4\left(2 \tan ^{-1} x\right)+2\left(2 \tan ^{-1} x\right)=\frac{\pi}{3} \\ &\Rightarrow 6 \tan ^{-1} x-8 \tan ^{-1} x+4 \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow 2 \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{6} \end{aligned}$ $\Rightarrow \tan ^{-1} x=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ $\left[\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\right]$ $\Rightarrow x=\frac{1}{\sqrt{3}}$ This is required solution.
Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 3.
Answer: $x=\frac{1}{\sqrt{3}}$ Given: $\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}, x>0$ Hint: Use
$\text { 1. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \text { and }$ $\text { 2. } \cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$ Solution: We have,
$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}$ We know that,
$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$ and convert
$\cot ^{-1}$ into
$\tan ^{-1}$ $\Rightarrow 2 \tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$ $\left[\cot \theta=\frac{1}{\tan \theta}\right]$ $\Rightarrow 2 \tan ^{-1} x+2 \tan ^{-1} x=\frac{2 \pi}{3}$ $\left[\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]$ $\begin{aligned} &\Rightarrow 4 \tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{2 \pi}{12} \\ &\Rightarrow x=\tan \frac{\pi}{6} \end{aligned}$ $\left[\tan ^{-1} x=\frac{\pi}{6} \Rightarrow x=\tan \frac{\pi}{6}\right]$ $\Rightarrow x=\frac{1}{\sqrt{3}}$ $\left[\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$ Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 4.
Answer: $\mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$ Given: $2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x), x \neq \frac{\pi}{2}$ Hint: Using
$2 \tan ^{-1}(x)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$ Solution: We have,
$2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x)$ We know that
$2 \tan ^{-1}(x)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$ $\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{2 \sin x}{1-\sin ^{2} x}\right)=\tan ^{-1}(2 \sec x) \\ &\Rightarrow \frac{2 \sin x}{\cos ^{2} x}=2 \sec x \end{aligned}$ $\left[1-\sin ^{2} x=\cos ^{2} x\right]$ $\Rightarrow \frac{\sin x}{\cos x \cos x}=\frac{1}{\cos x}$ $\left[\sec x=\frac{1}{\cos x}\right]$ $\Rightarrow \tan x=1$ $\left[\frac{\sin x}{\cos x}=\tan x\right]$ $\begin{aligned} &\Rightarrow x=\tan ^{-1}(1) \\ &\Rightarrow x=\frac{\pi}{4} \end{aligned}$ $\left[\tan ^{-1}(1)=\frac{\pi}{4}\right]$ Hence the principal value of
$x=n \pi+\frac{\pi}{4}, n \in Z$ Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 5.
Answer: $x=\sqrt{3}$ Given: $\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$ Hint: Using formula
$\text { 1. } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x$ $\text { 2. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$ Solution: We have,
$\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$ We know that,
$\begin{aligned} &\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \\ &\Rightarrow \cos ^{-1}\left[\frac{-\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \end{aligned}$ $[\cos (-\theta)=\pi-\cos \theta]$ $\Rightarrow \pi-\cos ^{-1}\left[\frac{\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$ $\left[\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]$ $\begin{aligned} &\Rightarrow \pi-2 \tan ^{-1}(x)+\tan ^{-1}(x)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-\tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\pi-\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow x=\tan \frac{\pi}{3} \end{aligned}$ $\Rightarrow x=\sqrt{3}$ $\left[\tan \left(\frac{x}{3}\right)=\sqrt{3}\right]$ Hence
$x=\sqrt{3}$ is required solution.
Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 6.
Answer: $x=\sqrt{3}$ Given:$\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$ Hint: Using formula
$\text { 1. } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x$ $\text { 2. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$ Solution: We have,
$\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$ We know that,
$\begin{aligned} &\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \\ &\Rightarrow \cos ^{-1}\left[\frac{-\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \end{aligned}$ $[\cos (-\theta)=\pi-\cos \theta]$ $\begin{aligned} &\Rightarrow \pi-\cos ^{-1}\left[\frac{\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-2 \tan ^{-1}(x)+\frac{1}{2} \times 2 \tan ^{-1} x=\frac{2 \pi}{3} \end{aligned}$ $\left[\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]$ $\begin{aligned} &\Rightarrow \pi-2 \tan ^{-1}(x)+\tan ^{-1}(x)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-\tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\pi-\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow x=\tan \frac{\pi}{3} \\ &\Rightarrow x=\sqrt{3} \end{aligned}$ $\left[\tan \left(\frac{x}{3}\right)=\sqrt{3}\right]$ Hence
$x=\sqrt{3}$ is required solution.
Inverse Trigonometric Function Exercise 3.14 Question 9.
Answer: $2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$ Given: $2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$ Hint: Using $2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ Solution: Let us assume$\begin{aligned} \text { L.H.S } &=2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right] \\ &=\cos ^{-1}\left[\frac{1-\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}{1+\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}\right] \end{aligned}$ $\left[2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]$ $=\cos ^{-1}\left[\frac{1-\left(\frac{a-b}{a+b}\right) \tan ^{2} \frac{\theta}{2}}{1+\left(\frac{a-b}{a+b}\right) \tan ^{2} \frac{\theta}{2}}\right]$ $\begin{aligned} &=\cos ^{-1}\left[\frac{a+b-(a-b) \tan ^{2} \frac{\theta}{2}}{a+b+(a-b) \tan ^{2} \frac{\theta}{2}}\right] \\ &=\cos ^{-1}\left[\frac{a\left(1-\tan ^{2} \frac{\theta}{2}\right)+b\left(1+\tan ^{2} \frac{\theta}{2}\right)}{a\left(1+\tan ^{2} \frac{\theta}{2}\right)+b\left(1-\tan ^{2} \frac{\theta}{2}\right)}\right] \end{aligned}$ Dividing numerator and denominator by $\left(1+\tan ^{2} \frac{\theta}{2}\right)$ we get$=\cos ^{-1}\left(\frac{a\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)+b}{a+b\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)}\right]$ $=\cos ^{-1}\left[\frac{a \cos \theta+b}{a+b \cos \theta}\right]$ $\left[\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}=\cos \theta\right]$ $= R.H.S$ Hence $2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$
Inverse Trigonometric Functions Exercise 3.14 Question 10.
Answer: $\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$ Given: $\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)$ where
$\alpha=a x-b y \: \: \& \: \: \beta=a y+b x$ Hint: Use
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Solution : Let us assume
L.H.S:
$=\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)$ We know that
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\begin{aligned} &=\tan ^{-1}\left[\frac{\frac{2 a b}{a^{2}-b^{2}}+\frac{2 x y}{x^{2}-y^{2}}}{1-\left(\frac{2 a b}{a^{2}-b^{2}}\right)\left(\frac{2 x y}{x^{2}-y^{2}}\right)}\right] \\ &=\tan ^{-1}\left[\frac{2 a b x^{2}-2 a b y^{2}+2 x y a^{2}-2 x y b^{2}}{a^{2} x^{2}-a^{2} y^{2}-b^{2} x^{2}+b^{2} y^{2}-4 a b x y}\right] \\ &=\tan ^{-1}\left[\frac{2\left(a b x^{2}+x y a^{2}-a b y^{2}-x y b^{2}\right)}{a^{2} x^{2}-a^{2} y^{2}-b^{2} x^{2}+b^{2} y^{2}-4 a b x y}\right] \\ &=\tan ^{-1}\left[\frac{2[a x(b x+a y)-b y(a y+b x)]}{(a x-b y)^{2}-\left(a^{2} y^{2}+b^{2} x^{2}+2 a b x y\right)}\right] \end{aligned}$ $=\tan ^{-1}\left[\frac{2(b x+a y)(a x-b y)}{(a x-b y)^{2}-(b x+a y)^{2}}\right]$ $\left[(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$ $=\tan ^{-1}\left[\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right]$ $[\alpha=a x-b y \& \beta=a y+b x]$ $=R.H.S$ Hence ,
$\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$ Inverse Trigonomeric Functions Exercise 3.14 Question 11.
Answer: $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$ Given : For any a, b, x, y > 0 We have to prove that $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$ where,$\alpha=-a x+b y \& \beta=b x+a y$ Hint: First we will divide numerator and denominator of first function and second function by $b^{3}$ and $y^{3}$ respectively, then use $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Solution: Let us assume$\begin{aligned} L . H . S &=\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right) \\ &=\frac{2}{3} \tan ^{-1}\left[\frac{\frac{3 a b^{2}-a^{3}}{b^{3}}}{\frac{b^{3}-3 a^{2} b}{b^{3}}}\right]+\frac{2}{3} \tan ^{-1}\left[\frac{\frac{3 x y^{2}-x^{3}}{y^{3}}}{\frac{y^{3}-3 x^{2} y}{y^{3}}}\right] \end{aligned}$ Dividing numerator and denominator of first function and second function by $b^{3}$ and $y^{3}$ respectively.$=\frac{2}{3} \tan ^{-1}\left[\frac{3\left(\frac{a}{b}\right)-\left(\frac{a}{b}\right)^{3}}{1-3\left(\frac{a}{b}\right)^{2}}\right]+\frac{2}{3} \tan ^{-1}\left[\frac{3\left(\frac{x}{y}\right)-\left(\frac{x}{y}\right)^{3}}{1-3\left(\frac{x}{y}\right)^{2}}\right]$ We know that,$3 \tan ^{-1} x=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$ $\Rightarrow \mathrm{LH} . \mathrm{S}=\frac{2}{3}\left[3 \tan ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\right]+\frac{2}{3}\left[3 \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\right]$ $\begin{aligned} &=2 \tan ^{-1}\left(\frac{a}{b}\right)+2 \tan ^{-1}\left(\frac{x}{y}\right) \\ &=2\left[\tan ^{-1}\left(\frac{a}{b}\right)+\tan ^{-1}\left(\frac{x}{y}\right)\right] \end{aligned}$ $=2 \tan ^{-1}\left[\frac{\frac{a}{b}+\frac{x}{y}}{1-\frac{a}{b} \times \frac{x}{y}}\right]$ $\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$ $\begin{aligned} &=2 \tan ^{-1}\left[\frac{a y+b x}{b y-a x}\right] \\ &=2 \tan ^{-1}\left(\frac{\beta}{\alpha}\right) \end{aligned}$ $[b y-a x=\alpha \& \beta=a y+b x]$ $=\tan ^{-1}\left(\frac{2 \times \frac{\beta}{\alpha}}{1-\left(\frac{\beta}{\alpha}\right)^{2}}\right)$ $\left[2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$ $\begin{aligned} &=\tan ^{-1}\left(\frac{2 \beta}{\alpha} \times \frac{\alpha^{2}}{\alpha^{2}-\beta^{2}}\right) \\ &=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right) \\ &=\text { R.H.S } \end{aligned}$ Hence,$\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$ Where $\alpha=-a x+b y \& \beta=b x+a y$
RD Sharma Class 12 Solutions Chapter 3 Exercise 3.14 involves inverse trigonometric problems related to cosecant, secant, cosine, tangent functions. Solving these problems will make a student grasp better and perform well in exams.
In calculus, Inverse Trigonometric Functions are used to find various integrals. In addition to mathematics, inverse trigonometric functions are applied in science and engineering. Students will learn about the domain and range restrictions of trigonometric functions that ensure the existence of their inverses and how to observe their behavior using graphical representations and examples.
The specific exercise 3.14 has 31 questions, including subparts, and now a student can practice a lot and gain knowledge about the topic. In addition, this exercise will help in overall understanding of the chapter. RD Sharma Class 12 Solutions Chapter 3 Exercise 3.14 has practice questions which are to be solved and are important for exam point of view. The concepts are now easily understood by the students.
Created by experts
With these solutions, the approach to learning and understanding topics will be forever altered. RD Sharma Solutions is here that has been created by a team of experts to ensure that every student excels in mathematics, which is why the book includes a variety of tricks and tips.
Best solutions for preparation
RD Sharma Class 12 textbook has high-level questions which sometimes become difficult for a student to solve. The RD Sharma Class 12 Solutions makes it easy. The chapters and exercises are covered wonderfully to help students in every possible way.
Students can refer to these solutions to gain extra knowledge about the topics. It is advisable to solve the practice questions to avail the benefit of these solutions.
NCERT based questions
Students can refer to these solutions to gain extra knowledge about the topics. It is advisable to solve the practice questions to avail the benefit of these solutions. The questions are set from NCERT books as teachers consider it as a standard.The students must buy RD Sharma Class 12 Solutions Chapter 1 Ex 1.2 to score well in board exams.
Different ways to solve a question
As the team of experts format the solutions, there are many ways which come out to solve one question. A student gets benefit from this technique. Here the illustrated examples are also included to make the learning easy and clarify the concepts a priority.
Free of cost
Career360 provides these solutions free of cost, and every student must refer to these solutions to get the benefit. Therefore, students have a golden opportunity to learn and get a lot of knowledge from these solutions. Many students have availed of the opportunity given by Career360, and now it is your turn to grab the exciting offer. If still unsure, you can also visit the website and find more expert-created answers.
Chapter-wise RD Sharma Class 12 Solutions